I have a question for Python if Moon.

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First of all, the reason why it doesn't work the way you want it is because you are not familiar with how to use the index method of the list.

The index returns the location in the list if the item exists, but an exception occurs if it does not exist.

try:                                      
    if a.index('%'):                      
       del a[a.index('%')]
    elif a.index(' '):                  
       del a[a.index(' ')]
except:                                  
    pass

If you put 1 space, 2 spaces, 3 spaces, and 3 spaces in the input...

a.index('%') does not have '%', so it will be excluded.

Of course elif does not work.

If index(''), returns only the location of the first found item.

This means that there are two spaces, but only one item is deleted.

To express what the questioner is trying to do, simply delete the blank or '%'.

In [17]: args = input()
1%2 3 4

In [18]: argsList = list(args)

In [19]: argsList
Out[19]: ['1', '%', '2', ' ', '3', ' ', '4']

In [20]: list(''.join(argsList).replace('%', '').replace(' ', ''))
Out[20]: ['1', '2', '3', '4']